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a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
\(A=x^2-16-6x-2x^2+x^2+6x+9=-7\\ B=\left(x^2+4\right)\left(x^2-4\right)-x^4+9\\ B=x^4-16-x^4+9=-7\)
a) \(A=\left(x+4\right)\left(x-4\right)-2x\left(3+x\right)+\left(x+3\right)^2\)
\(=x^2-16-2x^2-6x+x^2+6x+9=-7\)
b) \(B=\left(x^2+4\right)\left(x+2\right)\left(x-2\right)-\left(x^2+3\right)\left(x^2-3\right)\)
\(=\left(x^2+4\right)\left(x^2-4\right)-\left(x^4-9\right)\)
\(=x^4-16-x^4+9=-7\)
a: =>8x^2-20x+20x-50+4x(4x^2-12x+9)=0
=>8x^2-50+8x^3-48x^2+36x=0
=>8x^3-40x^2+36x-50=0
=>\(x\simeq4,29\)
b: =>(2x-3-3x-1)(2x-3+3x+1)=0
=>(-x-4)(5x-2)=0
=>x=2/5 hoặc x=-4
\(a,x\left(8x-2\right)-8x^2+12=0\)
\(\Rightarrow8x^2-2x-8x^2+12=0\)
\(\Rightarrow-2x+12=0\)
\(\Rightarrow-2x=-12\)
\(\Rightarrow x=6\)
\(b,x\left(4x-5\right)-\left(2x+1\right)^2=0\)
\(\Rightarrow4x^2-5x-4x^2-4x-1=0\)
\(\Rightarrow-9x-1=0\)
\(\Rightarrow-9x=1\)
\(\Rightarrow x=\frac{-1}{9}\)
a) x(8 - 2) - 8x2 + 12 = 0
x(8 - 2) - 8x2 = 12 - 0
x(8 - 2) - 8x2 = 12
2x = 12
x = 6
b) x(4x - 5) - (2x + 1)2 = 0
9x - 1 = 0
9x = 0 + 1
9x = 1
x = -1/9
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
a: \(\Leftrightarrow8x\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{0;3;-3\right\}\)
b: \(\Leftrightarrow x^2-4x+4-x^2-2x+3=12\)
=>-6x=5
hay x=-5/6